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% Author, Title, etc.

\title[]
{%
 A Brief Review Of 'Orientability and energy minimization 
in liquid crystal models'
%
}

\author[Ye ShiWei]
{
  Ye ShiWei\inst{1}
}

\institute[]
{
  \inst{1}%
  Peking University 
}

\date[2011]
{2011}



% The main document

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}{Outline}
  \tableofcontents
\end{frame}


\section{Over view}

\begin{frame}{What is the Paper about?}
  We(the authors) study the differences and the overlaps between the two theories:
\alert{Oseen-Frank theory} and \alert{Landau - de Gennes theory}.The two theories 
that modelled the uniaxial nematic liquid crystals\dots
\end{frame}

\begin{frame}[t]{The two theories}
  \begin{block}{Oseen-Frank theory}
    \begin{itemize}
    \item Uniaxial nematic liquid crystals are modelled through a unit vector field $n$
    \item It does not respect the head-to-tail symmetry.
    \end{itemize}
  \end{block}
  \begin{block}{Landau - de Gennes theory}
    \begin{itemize}
    \item Works with the tensor: $Q=s(n\otimes n - \frac{1}{3} Id)$
    \item The symmetry is preserved.
    \end{itemize}
  \end{block}
\end{frame}

\section{Introduction}

\begin{frame}[t]{What's done in the paper}
  \begin{block}{The works}
    \begin{itemize}
    \item Study the difference between the two theories.
    \item Establish when de Gennes can be replaced by Oseen-Frank, 
      and when this can not.
    \item<alert@1->
      Particularly, examine the instance when one would detect fake 'defects' the material resulting from
using the more simplistic Oseen-Frank model instead of the constrained Landau - de Gennes model 
    \end{itemize}
  \end{block}
  
\end{frame}

\begin{frame}[t]{How the de Gennes model goes}
  \begin{block}{Why tensor?}
    Because of the symmetry the first moment of the
probability measure vanishes.
  \end{block}

  \begin{block}{Q-tensor}
    \begin{itemize}
      \item $M_0=\frac{1}{4\pi } \intop_{S^2}p \otimes p dA=\frac{1}{3}Id$
      \item $Q=M-M_0=\intop_{S^2}(p \otimes p - \frac{1}{3} Id)d\mu (p)$
      \item In the restricted class of $Q \in W^{1,2}$,with $Q$ uniaxial,
        constrained Landau - de Gennes theory $Q=s(n \otimes n-\frac{1}{3}Id)$
      \item s(the scalar order parameter associated to the tensor Q):
        $-\frac{1}{2}\leq s \leq 1$, a measure of order.
      \item These Q-tensors can be interpreted as line fields
    \end{itemize}
  \end{block}
\end{frame}

\begin{frame}{More\dots}
  \begin{enumerate}
  \item Analysis from the point of view of energy minimization.

  \item When a line field can be 'oriented' into a vector field, 
without changing its regularity class.
  \item It is possible that a global energy minimizer is a line 
    field that is 'non-orientable'. Oseen-Frank theory would fail 
    to recognize a global minimizer.(necessary and sufficient conditions)
  \end{enumerate}
\end{frame}

\section{Preliminaries and notation}

\subsection{Basic notation}

\begin{frame}{About Q-tensor}
    \begin{block}{Proposition 1}
      The necessary and sufficient conditions for a $3 \times 3$ matrix to 
      have such a representation as  $Q=s(n\otimes n - \frac{1}{3} Id)$
    \end{block}
  \begin{itemize}
    \item The sets:
      \begin{displaymath}
        \mathcal{Q} = \{ Q\in \mathcal{M}^{3\times3};Q=s(n\otimes n -\frac{1}{3}Id) \text{ for some } n \in \mathbb{S}^2\}
      \end{displaymath}

      \begin{displaymath}
        \mathcal{Q}^2 = \{ Q\in \mathcal{Q}; Q = s((n_1,n_2,0)\otimes (n_1,n_2,0) - \frac{1}{3}Id)\}
      \end{displaymath}
    \item The operator:
      \begin{displaymath}
        P(n)=s(n\otimes n -\frac{1}{3}Id).
      \end{displaymath}
      P provides us with a way of 'unorienting' an $\mathbb{S}^2$ -valued vector field
  \end{itemize}
\end{frame}

\begin{frame}{Other notation}
  \begin{itemize}
    \item set in $\mathbb{R}^m$ as class $\mathit{C}^k$ 
    \item domain: open and connected set
    \item $W^{1,p}(M,N)$ and $W^{1,p}_{\varphi}(M,N)$
  \end{itemize}
\end{frame}

\subsection{Orientable}
\begin{frame}{Defination of orientable}
  \begin{block}{Definition 1}
    Let $\Omega \subset \mathbb{R}^d$be a domain. We say that $Q\in W^{1,p}(\Omega,\mathcal{Q})$
    is orientable if there exists an $n\in W^{1,p}(\Omega,\mathbb{S}^2)$such that $p(n) = Q$ $\mathcal{L}^d a.e.$ 
    Otherwise we call Q  non-orientable.
  \end{block}
\end{frame}

\begin{frame}{Proposition 2}
  \begin{block}{Proposition 2}
    Let $\Omega \subset \mathbb{R}^3$ be a domain. An orientable line 
    field $Q(x) \in W^{1,p}(\Omega, Q), 1 \leq p \leq \infty$, can have
    only two orientations. More precisely, if $n, m \in W^{1,p}(\Omega, \mathbb{S}^2 )$ 
    with $P (n) = P (m) = Q$ and $n \neq m$ $ a.e.$ we have that $m = -n$ $a.e. in \Omega$.
  \end{block}

  \begin{block}{Proof}
    \begin{itemize}
      \item $m=\tau n$ with $\tau^2(x)=1$ $a.e.$
      \item For a.e. $x_2,x_3$, $\tau(x)n(x)$ and $n(x)$ are absolutely continuous, so 
        $\tau(x)=\tau(x)n(x)\cdot\tau(x)$ is absolutely continuous in $x_1$.
      \item $\tau(x)$ is constant in $x_1$. With Fubini's theorem, we get $\tau_{,1}=0$
        and then $\tau_{,2}=0$, $\tau_{,3}=0$

    \end{itemize}
  \end{block}
\end{frame}

\begin{frame}{Proposition 3}
  \begin{block}{Proposition 3}
    Let $\Omega \subset \mathbb{S}^3$ be a bounded open set with Lipschitz boundary. 
    Let $Q \in W{1,p} (\Omega, Q)$ be orientable, so that $Q = s (n \otimes n - \frac{1}{3}Id)$ $ a.e.$ 
    in  for some $n \in W^{1,p} (\Omega, \mathbb{S}^2 )$. Then $Tr n \in L^p (\partial \Omega, \mathbb(S)^2 )$
    and Q is orientable on the boundary, i.e.
    \begin{displaymath}
      Tr Q = s \Big{(}Tr n \otimes Tr n - \frac{1}{3}Id\Big{)}
    \end{displaymath}
    with $Tr Q \in L^p (\partial \Omega, \mathcal{Q})$.
  \end{block}
  
  \begin{block}{Proof}
    \begin{itemize}
      \item extend,Truncate,Mollify
      \item Refer some results from \textit{Measure theory and fine properties of functions}.
    \end{itemize}
  \end{block}
\end{frame}

\begin{frame}{Proposition 4}
  \begin{block}{Proposition 4}
    Let $\Omega \subset \mathbb{R}^d$ be a bounded domain with boundary of class $C^0$. 
    For $1 \leq p \leq \infty$ let $Q^{(k)} \in W^{1,p} (\Omega, \mathcal{Q}), k \in \mathbb{N}$ 
    be a sequence of orientable maps with the corresponding $n^{(k)} \in W^{1,p} (\Omega, \mathbb{S}^2 )$
    such that $P (n^{(k)} ) = Q^{(k)}$ . If $Q^{(k)}$ converges weakly to $Q$ in $W^{1,p}$ , 
    where $1 \leq p < \infty$ (or weak* when $p = \infty$), then Q is orientable and $Q = P (n)$ 
    for some $n \in W^{1,p}$ .

  \end{block}
  
  \begin{block}{Proof}
    \begin{itemize}
      \item Lemma 1
      \item $n^{(k)}$ is bounded.subsequence $(n^{(k_l)} )$ $k_l \in \mathbb{N}$ we have $n^{(k_l)}\rightarrow 
n$ in $W^{1,p}(\Omega)$ and $n^{(k_l)} \rightarrow n$ $a.e.$ which implies that $P (n) = Q$.

    \end{itemize}
  \end{block}
\end{frame}

\begin{frame}{Lemma 1}
  \begin{block}{Lemma 1}
    Let $\Omega \in \mathbb{R}$ be an open and bounded set. 
    If $n \in W^{1,p}(\Omega, \mathbb{S}^2 ), 1 \leq p \leq \infty$ then $Q = P (n) \in
    W^{1,p} (\Omega, \mathcal{Q})$. Conversely, assume that $Q \in W^{1,p}(\Omega, \mathcal{Q}), 1 \leq p \leq \infty$
    and let n be a measurable function on $\Omega$
    with values in $\mathbb{S}^2$ such that $P (n) = Q$. Moreover assume that n is continuous along
    almost every line parallel to the coordinate axes and intersecting $\Omega$. Then $n \in W^{1,p} (\Omega, Q)$.
    Moreover:
    \begin{displaymath}
      Q_{ij,k} n_j = sn_{i,k }.
    \end{displaymath}
  \end{block}
\end{frame}

\begin{frame}{Lemma 2}
  \begin{block}{Lemma 2}
    Let $Q \in W^{1,p}(\Omega, \mathcal{Q}), 1 \leq p \leq \infty$ be non-orientable.
    Then there exists $\varepsilon > 0$, depending on Q, so
    that for all $\tilde{Q} \in W^{1,p}(\Omega, \mathcal{Q})$ with 
    $\lVert \tilde{Q} - Q\rVert_{ W^{1,p}(M,\mathbb{R}^9 )} < \varepsilon $ the line field Q is also non-orientable.

  \end{block}
  \begin{block}{Proof}
     Reasoning by contradiction
  \end{block}
\end{frame}

\begin{frame}{Elastic Enegy}
  \begin{block}{Elastic Enegy}
    The Oseen-Frank elastic energy is the same as the Landau - de Gennes elastic energy.
    For $Q(n)=s(n\otimes n-\frac{I}{3})$.
    \begin{itemize}
      \item The Oseen-Frank elastic energy:
        \begin{displaymath}
          E_{OF}=\int_{\Omega}W(n,\triangledown n) dx
        \end{displaymath}
      \item The Landau - de Gennes elastic energy:
        \begin{displaymath}
          \psi(Q, Q) = L_1 I_1 + L_2 I_2 + L_3 I_3 + L_4 I_4
        \end{displaymath}
    \end{itemize}
  \end{block}
\end{frame}

\section{Orientability issues}


\subsection{over view of this section}

\begin{frame}{over view of this section}
  \begin{block}{over view of this section}
    \begin{enumerate}
      \item We can define the map $b : Q \rightarrow \mathbb{R}P^2$, so that b is an isometry.
      \item P is a covering map.$P(n)=s(n\otimes n -\frac{Id}{3})$
      \item  one can check orientability of a {\color{red}{continuous}} line field $Q (on G)$
        just by determining the orientability of $Q|_{\partial G}$ 
      \item Functions in the Sobolev space $W^{1,p}(\Omega, Q)$ (with $\Omega \subset \mathbb{R}^d,d = 2, 3)$
        are not necessarily continuous for $p \leq d$.
      \item Also we are interested only in liftings that preserve the Sobolev regularity class.
    \end{enumerate}
  \end{block}
\end{frame}


\subsection{Continuous line fields on arbitrary domains}
\begin{frame}{Lemma 3(1)}
  \begin{block}{Lemma 3(1)}
    If $-\infty < t_1 < t_2 < \infty$ and $Q : [t1 , t2 ] \rightarrow \mathcal{Q}$ is 
    continuous then there exist exactly two {\color{red}{continuous}}
    maps (liftings) $n+ , n- : [t_1 , t_2 ] \rightarrow \mathbb{S}^2$ , so that
    \begin{displaymath}
      Q(t) = s(n^{\pm} (t) \otimes n^{\pm} (t) - \frac{Id}{3}),
    \end{displaymath}
    and $n^+ = -n^-$ . (Equivalently, given either of the two possible initial orientations 
    $\bar{n} \in \mathbb{S}^2$ , so that $Q(t1 ) = s(\bar{n}\otimes \bar{n} - \frac{Id}{3})$,
    there exists a unique continuous lifting $n : [t_1 , t_2 ] \rightarrow \mathbb{S}^2$ 
    with $n(t_1 ) = \bar{n}$.)
    \end{block}
\end{frame}

\begin{frame}{Lemma 3(2)}
  \begin{block}{Lemma 3(2)}
    Suppose in addition that $\bar{Q} = s(\bar{m}\otimes \bar{m} - \frac{Id}{3}), m \in \mathbb{S}^2$
    , and that $\lvert Q(t) - \bar{Q}\rvert \leq \varepsilon \lvert s \rvert$ 
    for all $t \in [t_1 , t_2 ]$ where
    $0 < \varepsilon < \sqrt{2}$. Then one of the liftings, $n^+$ say, satisfies
    \begin{displaymath}
      \lvert n^+ (t) - \bar{m}\rvert \leq \varepsilon, for\, All\, t \in [t_1 , t_2 ]
    \end{displaymath}
    and the other $n^- = -n^+$ satisfies
    \begin{displaymath}
      \lvert n^+ (t) + \bar{m}\rvert \leq \varepsilon, for\, All\, t \in [t_1 , t_2 ]
    \end{displaymath}
  \end{block}

\end{frame}

\begin{frame}{Proposition 5}
  \begin{block}{Proposition 5}
    Let $\Omega\subset \mathbb{R}^d$ be a simply-connected domain, and let $Q : \Omega \rightarrow \mathcal{Q}$
    be continuous. Then there   exists a {\color{red}{continuous}} lifting $n^Q : \Omega \rightarrow \mathbb{S}^2$ .
  \end{block}
In the rest of this subsection we restrict ourselves to a class of topologically non-trivial domains in 2D and
show that one can check orientability at the boundary.
  \begin{block}{Domain with holes}
    Let $\Omega \subset  \mathbb{R}^2$ be a bounded domain with $\partial \Omega$
    a Jordan curve. For $1 \leq i \leq N$ let $\omega_i \subset \mathbb{R}^2$ be a bounded
    domain with $\partial \omega_i$ a Jordan curve and $\omega_i \subset \Omega$, $\omega_i \cap \omega_j = \emptyset \, if\, i \neq j$.
    Note that $\Omega$ and $\omega_i$ are simply connected
    \begin{displaymath}
      G=\Omega \backslash \cup_{i=1}^{N}\bar{\omega_i}
    \end{displaymath}
  \end{block}
\end{frame}

\begin{frame}{Theorem 1}
  \begin{block}{Theorem 1}
    Let $Q : \bar{G} \rightarrow \mathcal{Q}_2$ be {\color{red}{continuous}} with $Q|_{\partial \omega_i}$
    orientable as a {\color{red}{continuous}} function for $1 \leq i \leq N $.
    Then $Q$ is orientable as a continuous function.
  \end{block}
  \begin{block}{Proof of theorem 1}
    \begin{itemize}
      \item Lemma 4
      \item Lemma 5
      \item {\color{red}{Lemma 6}}
      \item Lemma 7
    \end{itemize}
  \end{block}
\end{frame}

\begin{frame}{Lemma 6}
  \begin{block}{Lemma 6}
    Let $G$ be a domain with holes as above and let $Q \in C(\bar{G}, \mathcal{Q})$.
    There exists $\nu > 0$ such that if $\bar{x} \in G$,
    $ -\infty < t_1 < t_2 < \infty$, $f^{(j)} , f \in C([t_1 , t_2 ],\bar{G})$
    ,$ f^{(j)} ([t_1 , t_2 ]) \subset B( \bar{x}, \nu), f^{(j)} (t_2 ) \rightarrow f(t_2)$, and if
    $n^{(j)} , n: [t_1 , t_2 ] \rightarrow \mathbb{S}^2$ are continuous liftings of 
    $Q(f^{(j)} (\cdot)), Q(f (\cdot))$ respectively with $\lvert n^{(j)} (t_1 ) - n(t_1 )\rvert < 1$, then
    $n^{(j)} (t_2 ) \rightarrow n(t_2)$

  \end{block}
\end{frame}

\subsection{Arbitrary line fields on simply-connected domains}

\begin{frame}{Theorem 2}
  \begin{block}{Theorem 2}
    Let $\Omega$ be a bounded simply connected domain in $\mathbb{R}^d , d = 2, 3$,
    with continuous boundary. Let $Q \in W^{1,p} (\Omega, \mathcal{Q})$. If $p \geq 2$ 
    there exists a lifting $\varphi^Q \in W^{1,p} (\Omega, \mathbb{S}^2 )$ so that $P \circ \varphi^Q = Q$.\\
    Moreover we have the estimate
    \begin{displaymath}
      c_1\lVert \bigtriangledown Q \rVert _{L^p} \leq 
      \lVert \bigtriangledown \varphi^Q \rVert _{L^p} \leq
      c_2\lVert \bigtriangledown Q \rVert _{L^p}
    \end{displaymath}
    with $c_1 , c_2$ constants that depend only on p.\\
     For $p < 2$ there exist line fields for which there is no lifting.
  \end{block}
\end{frame}

\section{Analytic orientability criteria in 2D}

\subsection{Preliminary}

\begin{frame}{Preliminary}
  In this section we restrict ourselves to planar line fields, i.e. the domain $\Omega$ 
  is a subset of $\mathbb{R}^2$ and the line field takes values only in $\mathcal{Q}_2$
  , not in the whole of $\mathcal{Q}$.\\
  Try to obtain an analogue of the previous theorem 1 for less regular functions, in $W^{1,2} (G)$.
  \begin{block}{Claim}
    We claim that if the line field is non-orientable, as a continuous line field, then it is also non-orientable in
    $H^{1/2}(\partial G)$.
  \end{block}
\end{frame}

\begin{frame}{Some definitions}
  \begin{block}{auxiliary complex-valued map A(Q)}
    \begin{displaymath}
      A(Q)=\frac{2}{s}Q_{11}-\frac{1}{3}+\mathnormal{i} \frac{2}{s}Q_{12},A(Q)\in \mathbb{S}^1\subset \mathbb{C}.
    \end{displaymath}
  \end{block}
  if $Z(n) = n_1 +\mathnormal{i}n_2$ then $A(Q) = Z^2 (n)$.

  The auxiliary map allows one to associate to a planar line field an auxiliary unit-length vector field.
  
\end{frame}

\begin{frame}{Some definitions}

  \begin{block}{Degree of a continuous mapping}
    If f is a smooth map whose domain is a compact manifold and p is a regular value of f, consider the finite set
    \begin{displaymath}
      f^{-1}(p)=\{x_1,x_2,\dots,x_n\}.
    \end{displaymath}
    By p being a regular value, in a neighborhood of each $x_i$ the map f is a local diffeomorphism (it is a covering map).
    Diffeomorphisms can be either {\color{red}{orientation preserving}} or {\color{red}{orientation reversing}}
    . Let r be the number of points 
    $x_i$ at which $f$ is orientation preserving and s be the number at which f is orientation reversing. 
    When the domain of f is connected, the number $r - s$ is independent of the choice of p and one
    defines the degree of f to be $r - s$
  \end{block}
\end{frame}

\subsection{Important Propostions}
\begin{frame}{Proposition 6}
  \begin{block}{Proposition 6}
    Let $\Omega$ be a smooth, bounded domain in $\mathbb{R}^2$ and let
    $Q \in W^{ 1,2} (\Omega, \mathcal{Q}_2 )$. We denote by 
    $(\partial \Omega)_i , i = 1,\dots, k$ the connected components of the boundary.
    For any $i \in \{1, 2, . . . , k\}$ the function 
    $Tr Q|_{(\partial \Omega)_i} \in H^{ 1/2} ((\partial \Omega)_i , \mathcal{Q}_2 )$, 
    is orientable (in the space $H^{1/2}$ ) if and only if $deg(A(Tr Q), (\partial \Omega)_i ) \in 2\mathbb{Z}$.
    Moreover if there exists a unit-length vector field $n$ such that 
    $Z(n) = n_1 + \mathnormal{i}n_2 \in H^{ 1/2} ((\partial \Omega)_i , \mathbb{S}^1 )$ 
    and $P (n) = Tr Q a.e. on (\partial \Omega)_i$ then 
    $deg(n, (\partial \Omega)_i ) =\frac{1}{2} deg(A(Tr Q), (\partial \Omega)_i )$.
  \end{block}
\end{frame}

\begin{frame}{Proposition 7}
  \begin{block}{Propostion 7}
    Let $G$ be a planar domain with holes as defined previous. Assume moreover that
    $\partial G$ is smooth. Let $Q \in W^{1,2} (G, \mathcal{Q}^2)$. Then Q is orientable if and only if
    $deg(A(Tr Q|\partial \Omega ), \partial \Omega) \in 2\mathbb{Z}$, 
    $deg(A(Tr Q|\partial \omega_i ), \partial \omega_i ) \in 2\mathbb{Z}, i = 1, . . . , n.$

  \end{block}
\end{frame}


\subsection{The final lemmas of this section}
\begin{frame}{The final lemmas of this section(1)}
  \begin{block}{Lemma 10}
    Let $G$ be an open set in $\mathbb{R}^2$ . The function 
    $Q \in W^{ 1,2} (G; \mathcal{Q}_2 )\cap C(G; \mathcal{Q}_2 )$ is orientable as a function
    in $W^{1,2}$ if and only if it is orientable as a continuous function.
  \end{block}
\end{frame}

\begin{frame}{The final lemmas of this section(2)}
%  \begin{block}{Lemma 11}
  Lemma 11\\
    The line field $\tilde{Q}$ as in below on the domain $\Omega$ is not orientable in 
    $W^{ 1,2} (\Omega; \mathcal{Q}_2 )$ or in $C(\Omega; \mathcal{Q}_2 )$.
    \begin{displaymath}
      \tilde{Q}=s(\tilde{n}\otimes \tilde{n}-\frac{1}{3}Id)\in W^{1,2}(\tilde{\Omega},\mathcal{Q}_2)
    \end{displaymath}
    \begin{displaymath}
      \tilde{\Omega}=\{ (x,y)\in[-1,1]\times[-1,0],\sqrt{x^2+y^2}\geq \frac{1}{2}\}\cup
    \end{displaymath}
    \begin{displaymath}
            \{ (x,y);y\geq0,\frac{1}{2}\leq\sqrt{x^2+y^2}\leq1\}
    \end{displaymath}
    \begin{displaymath}
      \tilde{n}(x,y)=\left\{
        \begin{array}{ll}
          (0,1,0 )& if\, (x,y)\in([-1,1]\times[-1,0])\\
          (-\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}},0)&if\, y\geq0,\frac{1}{2}\leq\sqrt{x^2+y^2}\leq1\\
        \end{array}\right.
    \end{displaymath}

%  \end{block}
\end{frame}

\begin{frame}{Image}
  \begin{figure}[h!]\centering
    \includegraphics[width=8cm]{./img1.png}
    \caption{Image for Lemma 11}
  \end{figure}
\end{frame}

\section{Enegy minimizing}

\begin{frame}{Minimizers}
  The minimizing Q-harmonic maps versus minimizing harmonic maps in the plane.
  \begin{block}{Minimizers}
    \begin{displaymath}
      \mathcal{I}_{\delta}(Q)=\int_{M_{\delta}}\lvert \bigtriangledown Q(x)\rvert^2 dx
    \end{displaymath}

    \begin{displaymath}
      \mathcal{J}_{\delta}(n)=2s^2\int_{M_{\delta}}\lvert \bigtriangledown n(x)\rvert^2 dx
    \end{displaymath}
  \end{block}
 
\end{frame}

\begin{frame}{Image}
  \begin{figure}[h!]\centering
    \includegraphics[width=7cm]{./img2.png}
    \caption{A simple geometry with boundary
conditions that allow both orientable and non-orientable line fields.}
  \end{figure}
\end{frame}

\begin{frame}{Lemma 12}
  \begin{block}{Lemma 12}
   Let $\bar{n}_{\delta} \in W^{ 1,2} (M_{\delta} , \mathbb{S}^1 )$
   be any global energy minimizer of $\mathcal{J}_{\delta} (n) in W^{ 1,2} (M ; \mathbb{S}^1 )$
   (subject to tangent vector-field boundary conditions on the outer boundary, as in Fig. 3b).
   Let $\bar{Q}_{\delta} \in W^{ 1,2} (M_{\delta} ; \mathcal{Q}_2 )$ be
   any global energy minimizer of $\mathcal{I}_{\delta} (Q) in W^{ 1,2} (M_{\delta} ; \mathcal{Q}_2 )$
   (subject to tangent line-field boundary conditions on
   the outer boundary, as in Fig. 3a).
   There exists a $\delta_0 > 1$ so that for any $\delta > \delta_0$ we have
   $\mathcal{I}_{\delta} (\bar{Q}_{\delta} ) < \mathcal{J}_{\delta} ( \bar{n}_{\delta} )$.
   \end{block}
\end{frame}
\end{document}


